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# Class 12 RD Sharma Solutions – Chapter 22 Differential Equations – Exercise 22.9 | Set 1

• Last Updated : 18 Mar, 2021

### Question 1. x2dy + y(x + y)dy = 0

Solution:

We have,

x2dy + y(x + y)dy = 0

dy/dx = -y(x + y)/x2

It is a homogeneous equation,

So, put y = vx               (i)

On differentiating both sides w.r.t x,

dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = -vx(x + vx)/x2

v + x(dv/dx) = -v – v2

x(dv/dx) = -2v – v2 On integrating both sides,   log|v/(v + 2)|1/2 = -log|x/c|

v/(v + 2) = c2/x2 yx2 = (y + 2x)c2    (Where ‘c’ is integration constant)

### Question 2. (dy/dx) = (y – x)/(y + x)

Solution:

We have,

(dy/dx) = (y – x)/(y + x)

It is a homogeneous equation,

So, put y = vx               (i)

On differentiating both sides w.r.t x,

dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = (vx – x)/(vx + x)

v + x(dv/dx) = (v – 1)/(v + 1)

x(dv/dx) = (v – 1)/(v + 1) – v

x(dv/dx) = (v – 1 – v2 – v)/(v + 1)

x(dv/dx) = -(v2 + 1)/(v + 1) On integrating both sides, ∫vdv/(v2+1)+∫dv/(v2+1)=-∫(dx/x) (1/2)log|v2 + 1| + tan-1(v) = log(c/x)

log|(y2 + x2)/x2| + 2tan-1(y/x) = log(c/x)2

log(y2 + x2) – log(x)2 + 2tan-1(y/x) = log(c/x)2

log(y2 + x2) + 2tan-1(y/x) = 2log(c)  (Where ‘c’ is integration constant)

### Question 3. (dy/dx) = (y2 – x2)/2yx

Solution:

We have,

(dy/dx) = (y2 – x2)/2yx

It is a homogeneous equation,

So, put y = vx               (i)

On differentiating both sides w.r.t x,

dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = (v2x2 – x2)/2vx2

v + x(dv/dx) = (v2 – 1)/2v

x(dv/dx) = [(v2 – 1)/2v] – v

x(dv/dx) = (v2 – 1 – 2v2)/2v

x(dv/dx) = -(v2 + 1)/2v On integrating both sides, log|v2+1| = -log(x) + log(c)

log|v2+1| = log(c/x)

y2/x2 + 1 = |c/x|

(x2 + y2) = cx  (Where ‘c’ is integration constant)

### Question 4. x(dy/dx) = (x + y)

Solution:

We have,

x(dy/dx) = (x+y)

(dy/dx) = (x+y)/x

It is a homogeneous equation,

So, put y = vx               (i)

On differentiating both sides w.r.t x,

dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = (x + vx)/x

v + x(dv/dx) = (1 + v)

x(dv/dx) = 1

dv = (dx/x)

On integrating both sides,

∫dv = ∫(dx/x)

v = log(x) + c

y/x = log(x) + c

y = xlog(x) + cx  (Where ‘c’ is integration constant)

### Question 5. (x2 – y2)dx – 2xydy = 0

Solution:

We have,

(x2 – y2)dx – 2xydy = 0

(dy/dx) = (x2 – y2)/2xy

It is a homogeneous equation,

So, put y = vx               (i)

On differentiating both sides w.r.t x,

dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = (x2 – v2x2)/2xvx

v + x(dv/dx) = (1 – v2)/2v

x(dv/dx) = [(1 – v2)/2v] – v

x(dv/dx) = (1 – 3v2)/2v On integrating both sides,  -(1/3)log(1 – 3v2) = log(x) – log(c)

log(1 – 3v2) = -log(x)3 + log(c) (x2 – 3y2)/x2 = (c/x3)

x(x2 – 3y2) = c  (Where ‘c’ is integration constant)

### Question 6. (dy/dx) = (x + y)/(x – y)

Solution:

We have,

(dy/dx) = (x + y)/(x – y)

It is a homogeneous equation,

So, put y = vx               (i)

On differentiating both sides w.r.t x,

dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = (x + vx)/(x – vx)

v + x(dv/dx) = (1 + v)/(1 – v)

x(dv/dx) = [(1 + v)/(1 – v)] – v

x(dv/dx) = (1 + v – v + v2)/(1 – v)

x(dv/dx) = (1 + v2)/(1 – v) On integrating both sides, ∫dv/(v2 + 1) – ∫vdv/(v2 + 1) = ∫(dx/x)

tan-1(v) – (1/2)log(v2 + 1) = log(x) + c

tan-1(y/x) – (1/2)log(y2/x2 + 1) = log(x) + c

tan-1(y/x) – (1/2)log(y2 + x2) + log(x) = log(x) + c

tan-1(y/x) = (1/2)log(y2 + x2) + c  (Where ‘c’ is integration constant)

### Question 7. 2xy(dy/dx) = (x2 + y2)

Solution:

We have,

2xy(dy/dx) = (x2 + y2)

(dy/dx) = (x2 + y2)/2xy

It is a homogeneous equation,

So, put y = vx               (i)

On differentiating both sides w.r.t x,

dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = (x2 + v2x2)/2xvx

v + x(dv/dx) = (1 + v2)/2v

x(dv/dx) = [(1 + v2)/2v] – v

x(dv/dx) = (1 – v2)/2v On integrating both sides, -log(1 – v2) = log(x) – log(c)

log(1 – v2) = -log(x) + log(c)

1 – y2/x2 = (c/x)

(x2 – y2) = cx  (Where ‘c’ is integration constant)

### Question 8. x2(dy/dx) = x2 – 2y2 + xy

Solution:

We have,

x2(dy/dx) = x2 – 2y2 + xy

(dy/dx) = (x2 – 2y2 + xy)/x2

It is a homogeneous equation,

So, put y = vx               (i)

On differentiating both sides w.r.t x,

dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = (x2 – 2v2x2 + xvx)/2xvx

v + x(dv/dx) = (1 – 2v2 + v)/x2

x(dv/dx) = (1 – 2v2 + v) – v

x(dv/dx) = (1 – 2v2)

dv/(1 – 2v2) = (dx/x)

On integrating both sides,

dv/(1 – 2v2) = ∫(dx/x)       (Where ‘c’ is integration constant)

### Question 9. xy(dy/dx) = x2 – y2

Solution:

We have,

xy(dy/dx) = x2 – y2

(dy/dx) = (x2 – y2)/xy

It is a homogeneous equation,

So, put y = vx               (i)

On differentiating both sides w.r.t x,

dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = (x2 – v2x2)/xvx

v + x(dv/dx) = (1 – v2)/v

x(dv/dx) = [(1 – v2)/v] – v

x(dv/dx) = (1 – 2v2)/v

vdv/(1 – 2v2) = (dx/x)

On integrating both sides,

∫vdv/(1 – 2v2) = ∫(dx/x)

∫4vdv/(1 – 2v2) = 4∫(dx/x)

-log(1 – 2v2) = 4log(x) – log(c)

log(1 – 2v2) = log(c/x4)

(1 – 2y2/x2) = c/x4

(x2-2y2)/x2 = c/x4

x2(x2 – 2y2) = c  (Where ‘c’ is integration constant)

### Question 10. yex/ydx = (xex/y + y)dy

Solution:

We have,

yex/ydx = (xex/y + y)dy

(dy/dx) = (xex/y + y)/yex/y

It is a homogeneous equation,

So, put x = vy               (i)

On differentiating both sides w.r.t x,

dx/dy = v + y(dv/dy)

So,

v + y(dv/dy) = (vyevy/y + y)/yevy/y

v + y(dv/dy) = (vev + 1)/ev

y(dv/dy) = [(vev + 1)/ev] – v

y(dv/dy) = (vev + 1 – vev)/ev

y(dv/dy) = (1/ev)

evdv = (dy/y)

On integrating both sides,

∫evdv = ∫(dy/y)

ev = log(y) + log(c)

ex/y = log(y) + log(c)  (Where ‘c’ is integration constant)

### Question 11. x2(dy/dx) = x2 + xy + y2

Solution:

We have,

x2(dy/dx) = x2 + xy + y2

dy/dx = (x2 + xy + y2)/x2

It is a homogeneous equation,

So, put y = vx               (i)

On differentiating both sides w.r.t x,

dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = (x2 + xvx + v2x2)/x2

v + x(dv/dx) = (1 + v + v2)

x(dv/dx) = (1 + v + v2) – v

dv/(1 + v2) = (dx/x)

On integrating both sides,

∫dv/(1 + v2) = ∫(dx/x)

tan-1(v) = log|x| + c

tan-1(y/x) = log|x| + c    (Where ‘c’ is integration constant)

### Question 12. (y2 – 2xy)dx = (x2 – 2xy)dy

Solution:

We have,

(y2 – 2xy)dx = (x2 – 2xy)dy

(dy/dx) = (y2 – 2xy)/(x2 – 2xy)

It is a homogeneous equation,

So, put y = vx               (i)

On differentiating both sides w.r.t x,

dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = (v2x2 – 2xvx)/(x2 – 2xvx)

v + x(dv/dx) = (v2 – 2v)/(1 – 2v)

x(dv/dx) = [(v2 – 2v – v + 2v2)/(1 – 2v)]

x(dv/dx) = 3(v2 – 1)/(1 – 2v)

-(2v – 1)dv/(v2 – v) = 3(dx/x)

On integrating both sides,

-∫(2v – 1)dv/(v2 – v) = 3∫(dx/x)

-log|v2 – v| = 3log|x| – log|c|

log|v2 – v| = log|c/x3|

(y2/x2 – y/x) = (c/x3)

(y2 – xy) = c/x

x(y2 – xy) = c (Where ‘c’ is integration constant)

### Question 13. 2xydx + (x2 + 2y2)dy = 0

Solution:

We have,

2xydx + (x2 + 2y2)dy = 0

dy/dx = -(2xy)/(x2 + 2y2)

It is a homogeneous equation,

So, put y = vx               (i)

On differentiating both sides w.r.t x,

dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = -(2xvx)/(x2 + 2v2x2)

v + x(dv/dx) = -(2v)/(1 + 2v2)

x(dv/dx) = -[(2v)/(1 + 2v2)] – v  On integrating both sides, Substituting (3v + 2v3) = z

On differentiating both sides w.r.t x,

3(1 + 2v)dv = dz

(1 + 2v)dv = (dz/3)

(1/3)∫(dz/z) = -∫(dx/x)

(1/3)log|z| = -log|x| + log|c|

log|3v + 2v3| = log|c/x|3

3y/x + 2(y/x)3 = (c/x)3

(3yx2 + 2y3) = c (Where ‘c’ is integration constant)

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